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Pascals Triangle

  1. Sep 7, 2009 #1
    how do i find the independent of x in (6x^(2)-3/x)^6

    I know that the formula nCr a^(n-r) b^r

    i have got an answer of n = 6 and r = 4 i just wanna make sure that im correct and i would like someone to give me a hand~ :)
     
  2. jcsd
  3. Sep 7, 2009 #2
    I don't think people (including myself) understand what you mean by finding "the independent of x".
     
  4. Sep 7, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Do you mean possibly the "constant term" (which is "independent of x") when you expand the polynomial? If so, then yes, r= 4 is correct. However, in future, it would be good to show how you got your answer so (just in case you are wrong!) we could indicate where you went wrong.
     
  5. Sep 7, 2009 #4
    I believe the sought expression is "coefficient". As in: Find the coefficient of x in the expansion of the expression [itex](6x^2-\frac{3}{x})^6[/itex].

    Obviously n = 6 here. In this case one would need to find what value of r results in [itex]a^r b^{(n-r)}[/itex] being a multiple of x (here [itex]a=6x^2,b=-\frac{3}{x}[/itex].

    --Elucidus
     
  6. Sep 9, 2009 #5
    independent of x simply means that x^0 and yes I am trying to find the co-efficient there

    heres are my working out.
    independent of x in (6x^(2)-3/x)^6

    1. (6^(6-r) x^(12-2r) ((-3^(r))/x^(r))

    2. x^(12-2r)/x^(r) = x^(0)

    3. x^(12-3r)=x^(0)

    4. 12-3r=0

    5. r = 4
     
  7. Sep 10, 2009 #6
    Ah, I see. This is a term I was unfamiliar with, but I understand now. Your work seems correct so far. If r = 4, what would the independent coefficient (constant term) be?

    --Elucidus
     
  8. Sep 12, 2009 #7
    thx for the help :)

    exams finally over
     
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